# log (2x) - log (x+4) = log 5

hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted on

log 2x - log (x+4) = log 5

We know that log a - log b = log a/b

==> log 2x/(x+4) = log5

==? 2x/(x+4) = 5

Multiply by (x+4):

==> 2x = 5*(x+4)

==> 2x = 5x + 20

==> 3x = -20

==> x = -20/3

jeyaram | Student, Undergraduate | (Level 1) Valedictorian

Posted on

log (2x) - log (x+4) = log 5

log(2x/[x+4])=log5

2x/(x+4)=5

2x=5(x+4)

2x=5x+20

3x=(-20)

x=(-20/3)

x=(-6.66666)

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

We'll impose the constraints of existence of logarithms:

2x>0

x>0

x+4>0

x>-4

The common interval of values that satisfies both constraints is (-0 , +inf.).

Now, we'll solve the equation. First, we'll add log (x+4) both sides:

log (2x) - log (x+4) + log (x+4) = log 5 + log (x+4)

log (2x) = log 5 + log (x+4)

Now, we'll use the product property of the logarithms:

log (2x) = log 5*(x+4)

Because the logarithms have matching bases, we'll use the one to one property:

2x = 5x + 20

We'll subtract 5x both sides:

2x-5x = 20

-3x = 20

We'll divide by -3:

x = -20/3 < 0

Since the solution is negative, is not admissible, so the equation has no solutions!